Oscillation Problem: Find Period and Potential Energy

[Misheel]

Homework Statement

mass "m" object is located near the origin of the coordinate system. External force was exerted on the object depending on the coordinate by the formula of Fx=-4*sin(3*pi*x). Find the short oscillation period, and the potential energy depending on the coordinate.

Homework Equations

F=ma

The Attempt at a Solution

I am not really sure what is the short oscillation period... but
since there is only 1 force:
F=Fx=ma
-4sin(3*pi*x)=m*(d^2*x)/dt^2

and assuming that the object will move very little (because it's said to be SHORT osccilation period ?)
sin(3*pi*x) is approximately 3*pi*x .
and since
-ω²*x=(d^2*x)/dt^2 is the formula of the harmonic oscillation :
we find ω=12*pi/m
and the T=2*pi*√(m/(12*pi)) ?

and also assuming that in SHORT oscillation :
it is almost like a spring :
F=-12*pi*x=-kx
k=12*pi

potential energy is Ep= k*x^2/2=6*pi*x^2 ?

 

[vela]

Homework Statement

mass "m" object is located near the origin of the coordinate system. External force was exerted on the object depending on the coordinate by the formula of Fx=-4*sin(3*pi*x). Find the short oscillation period, and the potential energy depending on the coordinate.

Homework Equations

F=ma

The Attempt at a Solution

I am not really sure what is the short oscillation period... but
since there is only 1 force:
F=Fx=ma
-4sin(3*pi*x)=m*(d^2*x)/dt^2

and assuming that the object will move very little (because it's said to be SHORT osccilation period ?)
sin(3*pi*x) is approximately 3*pi*x .
and since
-ω²*x=(d^2*x)/dt^2 is the formula of the harmonic oscillation :
we find ω=12*pi/m
and the T=2*pi*√(m/(12*pi)) ?

This is fine.

and also assuming that in SHORT oscillation :
it is almost like a spring :
F=-12*pi*x=-kx
k=12*pi

potential energy is Ep= k*x^2/2=6*pi*x^2 ?

I think the problem is looking for the potential corresponding to the original force, not the approximation.

 

[Misheel]

Thank You for your reply, Vela

umm...then, i have no other ideas other than using FORCE to solve this problem :PP which is :
F=-kx
Fx=F=-4*sin(3*pi*x)=-kx from this
we find k=4*sin(3*pi*x)/x
so Ep=kx^2/2=2*sin(3*pi*x)*x ?

is it right ? :P

Thanks

 

[vela]

How are potential energy and force related in general?

 

[Misheel]

How are potential energy and force related in general?

Maybe the work done by force is divided into object's konetik and potential energy ?